//给定二叉树的根节点 root ，返回所有左叶子之和。 
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// 示例 1： 
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//输入: root = [3,9,20,null,null,15,7] 
//输出: 24 
//解释: 在这个二叉树中，有两个左叶子，分别是 9 和 15，所以返回 24
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// 示例 2: 
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//输入: root = [1]
//输出: 0
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// 提示: 
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// 节点数在 [1, 1000] 范围内 
// -1000 <= Node.val <= 1000 
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// Related Topics 树 深度优先搜索 广度优先搜索 二叉树 👍 620 👎 0


package leetcode.editor.cn;

// [404]左叶子之和

public class SumOfLeftLeaves_404 {
    private class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;

        TreeNode() {
        }

        TreeNode(int val) {
            this.val = val;
        }

        TreeNode(int val, TreeNode left, TreeNode right) {
            this.val = val;
            this.left = left;
            this.right = right;
        }
    }

    public static void main(String[] args) {
        Solution solution = new SumOfLeftLeaves_404().new Solution();
    }
    //leetcode submit region begin(Prohibit modification and deletion)

    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     * int val;
     * TreeNode left;
     * TreeNode right;
     * TreeNode() {}
     * TreeNode(int val) { this.val = val; }
     * TreeNode(int val, TreeNode left, TreeNode right) {
     * this.val = val;
     * this.left = left;
     * this.right = right;
     * }
     * }
     */
    class Solution {
        int sum = 0;

        public int sumOfLeftLeaves(TreeNode root) {
            if (root == null) {
                return 0;
            }
            process(root);
            return sum;
        }

        public void process(TreeNode node) {
            if (node.left != null && node.left.left == null && node.left.right == null) {
                sum += node.left.val;
            } else if (node.left != null) {
                process(node.left);
            }
            if (node.right != null) {
                process(node.right);
            }
        }
    }
//leetcode submit region end(Prohibit modification and deletion)

}